Ternary scale theorems: Difference between revisions

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First we handle the ''a'' = 2''k'' case. This means that ''k'' = gcd(''a'', ''k'') = 1, and ''a'' = 2. This is the trivial ''n'' = 4 ('''xyxz''') case. Thus it remains to handle the cases (1) and (2) above. The steps of a MOS are distributed maximally evenly. Hence the positions occupied by '''x''' + '''X''' are a maximally even subset of <math>\mathbb{Z}/(n/2)\mathbb{Z},</math> and so are the positions occupied by 2'''X''' resp. 2'''x'''. Whenever the letter '''x''' + '''X''' is encountered, the number of the last letters that are equated to '''X''' that are consumed is 1, which is odd. Whenever the other letter is encountered, that number is even (0 or 2). Hence the letter 2'''X''' resp. 2'''x''' serves as the non-slot letter, and the letters ('''x''' + '''X''') serve as the slot letters where a 2-period filling MOS word (a repetition of ('''x'''+'''y''')('''x'''+'''z''') is substituted.

Now we count the letters that occur in these MOS substitution words of 2-steps. Consider the chunk boundaries of the template MOS. For every boundary between chunks, there is one slot letter in the template MOS for ''s''''1'' and one in the template MOS ''s''''2'', due to parity. So it suffices that we have evenly many boundaries between (nonempty) chunks. Equivalently, we have to prove that there are evenly many steps in the template MOS ''a'''''x''' 2''k'''''X''', which is true by assumption (''a'' is even, 2''k'' is even).

Now we count the letters that occur in these MOS substitution words of 2-steps. Consider the chunk boundaries of the template MOS. For every boundary between chunks, there is one slot letter in the template MOS for ''s''''1'' and one in the template MOS ''s''''2'', due to index parity. So it suffices that we have evenly many boundaries between (nonempty) chunks. Equivalently, we have to prove that there are evenly many steps in the template MOS ''a'''''x''' 2''k'''''X''', which is true by assumption (''a'' is even, 2''k'' is even).

* In the singly even case, since there are evenly many slot letters in both ''s''1 and ''s''2, there are oddly many non-slot letters in both. Since ''s''1 and ''s''2 differ by interchanging '''y''' and '''z''', they have "opposite" filling letters, '''x''' + '''y''' being the opposite of '''x''' + '''z'''. This makes ''s''1 and ''s''2 opposite chiralities of an odd-regular MV3 scale.

* In the doubly even case, the number of non-slot letters in ''s''1 and ''s''2 is even, and we have a filling MOS of period 2. Since ''s''1 and ''s''2 are both primitive, they are both even-regular scales. {{Qed}}

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 First we handle the ''a'' = 2''k'' case. This means that ''k'' = gcd(''a'', ''k'') = 1, and ''a'' = 2. This is the trivial ''n'' = 4 ('''xyxz''') case. Thus it remains to handle the cases (1) and (2) above. The steps of a MOS are distributed maximally evenly. Hence the positions occupied by '''x''' + '''X''' are a maximally even subset of <math>\mathbb{Z}/(n/2)\mathbb{Z},</math> and so are the positions occupied by 2'''X''' resp. 2'''x'''. Whenever the letter '''x''' + '''X''' is encountered, the number of the last letters that are equated to '''X''' that are consumed is 1, which is odd. Whenever the other letter is encountered, that number is even (0 or 2). Hence the letter 2'''X''' resp. 2'''x''' serves as the non-slot letter, and the letters ('''x''' + '''X''') serve as the slot letters where a 2-period filling MOS word (a repetition of ('''x'''+'''y''')('''x'''+'''z''') is substituted.
-Now we count the letters that occur in these MOS substitution words of 2-steps. Consider the chunk boundaries of the template MOS. For every boundary between chunks, there is one slot letter in the template MOS for ''s''<sub>''1''</sub> and one in the template MOS ''s''<sub>''2''</sub>, due to parity. So it suffices that we have evenly many boundaries between (nonempty) chunks. Equivalently, we have to prove that there are evenly many steps in the template MOS ''a'''''x''' 2''k'''''X''', which is true by assumption (''a'' is even, 2''k'' is even).
+Now we count the letters that occur in these MOS substitution words of 2-steps. Consider the chunk boundaries of the template MOS. For every boundary between chunks, there is one slot letter in the template MOS for ''s''<sub>''1''</sub> and one in the template MOS ''s''<sub>''2''</sub>, due to index parity. So it suffices that we have evenly many boundaries between (nonempty) chunks. Equivalently, we have to prove that there are evenly many steps in the template MOS ''a'''''x''' 2''k'''''X''', which is true by assumption (''a'' is even, 2''k'' is even).
 * In the singly even case, since there are evenly many slot letters in both ''s''<sub>1</sub> and ''s''<sub>2</sub>, there are oddly many non-slot letters in both. Since ''s''<sub>1</sub> and ''s''<sub>2</sub> differ by interchanging '''y''' and '''z''', they have "opposite" filling letters, '''x''' + '''y''' being the opposite of '''x''' + '''z'''. This makes ''s''<sub>1</sub> and ''s''<sub>2</sub> opposite chiralities of an odd-regular MV3 scale.
 * In the doubly even case, the number of non-slot letters in ''s''<sub>1</sub> and ''s''<sub>2</sub> is even, and we have a filling MOS of period 2. Since ''s''<sub>1</sub> and ''s''<sub>2</sub> are both primitive, they are both even-regular scales. {{Qed}}